# Show that bilinear form is symplectic

Let $$U$$ be a finite-dimensional $$\mathbb{R}$$-vector space. I want to show that the following bilinear form over $$V:=U\oplus U^\ast$$ is symplectic: $$\omega:V\times V\rightarrow\mathbb{R},\quad\omega((u,\varphi),(v,\Psi)):=\varphi(v)-\Psi(u)$$.

A symplectic form is a skew-symmetric bilinear form with trivial kernel. Bilinearity is obvious. From $$\varphi(v)-\Psi(u)=-(\Psi(u)-\varphi(v))$$, we see that $$\omega$$ is skew-symmetric.

Now to the kernel. We must show that $$\mathrm{ker}(\omega)=\{y\in V: \omega(x,y)=0\}=\{0\}$$ for all $$x\in V$$. So what I did is to see when $$\omega_x:=\omega(x,\cdot)=0$$ for a given $$x\in V$$. I find that for $$x:=(v,\varphi),y:=(u,\Psi)$$ we must have $$\varphi(v)=\Psi(u)$$ for all $$\varphi\in U^\ast,v\in U$$, i.e. $$\mathrm{ker}(\omega)=\{(u,\varphi)\times(v,\Psi)\in V:\varphi=\Psi\}\neq\{0\}$$. However, my lecture notes say that $$\omega$$ should have a trivial kernel.

Can anybody tell me where I went wrong?

I think the main issue here might be in the correct phrasing of $$\ker \omega$$. This link should provide an answer to your question Show skew-symmetric, non-degenerate bilinear form $((a, \varphi),(b, \psi)) \mapsto \langle(a, \varphi),(b, \psi) \rangle := \varphi(b)-\psi(a)$